• Problem:
A rectangular garden has an area of 84 m² and a perimeter of 38 m. Find its length and width.
• Solution:
To find the measure of its length and width, let's make two equations first for the problem. For the area, length times width while for the perimeter, twice the sum of length and width.
[tex] \large \tt l \times w = 84 [/tex]
[tex] \large \tt 2l + 2w = 38 [/tex]
Then let's simplify the second equation and try to find the measure of the length.
[tex] \frac{ \large \tt 2l + 2w = 38 }{\large \tt2} \\ \large\tt l + w = 19 \\ \large \tt l = 19 - w[/tex]
Let's substitute the initial value of length to the first equation.
[tex] \large \tt (19 - w)w = 84 \\ \large \tt 19w - w {}^{2} = 84 \\ \tt \large \tt w {}^{2} -19w + 84 = 0 \\ \large \tt w {}^{2} - 7w - 12w + 84 = 0 \\ \large \tt (w {}^{2} - 7w )- (12w - 84) = 0 \\ \large \tt w(w - 7) - 12(w - 7) = 0 \\ \large \tt (w - 12)(w - 7) = 0 \\ \overline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large \tt w - 12 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \large \blue{\tt w = 12 }\\ \large \tt w - 7 = 0 \\ \large \blue{\tt w = 7}[/tex]
Since length should measure longer than the width, then our width will be 7 meters.
Substitute the measure of width which is 7 to any of the equation to get the measure of the length.
[tex] \large \tt l = 19 - w \\ \large \tt l = 19 - 7 \\ \blue{ \large \tt l = 12}[/tex]
Thus, our length is 12 meters.
• Answer:
[tex] \large \boxed{ \tt C. \: I = 12 \: m, w = 7m }[/tex]
